Question: A curve in the plane is defined parametrically by the equations $x=\sqrt{1-3t}$ and $y=\dfrac{3}{t}$. Find the value of $\dfrac{dy}{dx}$ at $t=-1$. Choose 1 answer: Choose 1 answer: (Choice A) A $-12$ (Choice B) B $2$ (Choice C) C $-3$ (Choice D) D $4$
Explanation: In general, to find the derivative (i.e. the expression for $\dfrac{dy}{dx}$ ) of a function defined parametrically by the equations $x=u(t)$ and $y=v(t)$ (where $u$ and $v$ are any functions of $t$ ), we use the following rule: $\dfrac{dy}{dx}=\dfrac{\left(\dfrac{dy}{dt}\right)}{\left(\dfrac{dx}{dt}\right)}=\dfrac{v'(t)}{u'(t)}$ We are given that $x=\sqrt{1-3t}$ and $y=\dfrac{3}{t}$ : $\begin{aligned} \dfrac{dy}{dx}&=\dfrac{\dfrac{d}{dt}\left(\dfrac{3}{t}\right)}{\dfrac{d}{dt}(\sqrt{1-3t})} \\\\ &=\dfrac{\left(-\dfrac{3}{t^2}\right)}{\left(-\dfrac{3}{2\sqrt{1-3t}}\right)} \\\\ &=\dfrac{2\sqrt{1-3t}}{t^2} \gray{\text{Simplify}} \end{aligned}$ Now let's evaluate $\dfrac{dy}{dx}$ at $t= {-1}$ : $\begin{aligned} &\phantom{=}\dfrac{2\sqrt{1-3({-1})}}{({-1})^2} \\\\ &=\dfrac{2\cdot 2}{1} \\\\ &=4 \end{aligned}$ In conclusion, the value of $\dfrac{dy}{dx}$ at $t=-1$ is $4$.